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5
Linear Equations
5.1
Solution of a linear Equation with one Variable
A linear equation contains no indices greater than 1 of the variable x (say), e.g. 3x − 5 = 7
is a linear equation but 3x2 − 5 = 7 is a non-linear equation.
A linear equation can be represented pictorially as a straight line – see Topic 6.
The two sides of an equation are like the two sides of a set of scales in equilibrium.
Whatever you do to one side you must do to the other side to preserve the equilibrium.
L.H.S.
=
R.H.S.
Examples:
(a) Solve 3x − 5 = 7
3x − 5 + 5 = 7 + 5
Solution:
adding 5 to each side
3x = 12
3x
12
=
3
3
x = 4
simplifying
dividing both sides by 3
simplifying
Check by substituting x = 4 in the original equation
L.H.S. = 3 × 4 − 5 = 12 − 5
i.e.
= 7
=
∴
R.H.S.
x = 4
5–1
is the correct solution.
(b) Solve 4(x − 1) + 2 = 2x − 5
Solution
4x − 4 + 2 = 2x − 5
4x − 2 = 2x − 5
4x − 2 + 2 = 2x − 5 + 2
4x = 2x − 3
4x − 2x = 2x − 3 − 2x
2x = −3
3
2x
= −
2
2
1
x = −1
2
Check:
1
L.H.S. = 4 −1 − 1 + 2
2
1
= 4 −2
+2
2
= −10 + 2
R.H.S.
= −8
1
= 2 −1
−5
2
= −3 − 5
when x = −1
when x = −1
1
2
1
2
= −8
∴
= L.H.S.
1
x = −1
is the correct solution.
2
Note: While checking is seldom required in a problem, it is a useful exercise especially if
you are asked to use the answer in some further problem. So, make a habit of checking.
5–2
Exercises 5.1:
Solve
(i) 3x − 1 = 5
(ii) 4x + 2 = 9
(iii) 6 − 2x = 1
(iv) 5x + 1 = −9
(v) 3x + 1 = x − 5
(vi) 6x − 1 = 3 − 2x
(vii) 2 − 3(x − 2) = x + 4
(viii) 5(x − 1) + 1 = 2(x − 2).
In general, a linear equation in one variable has just one solution,
i.e., a unique solution.
[Examples of linear equations in one variable which do not have a unique solution are
4
3 x+
= 3x + 4 and 3x + 4 = 3(x + 4).
3
In the first equation, both sides of the equation are identical and so any value of x will
satisfy the equation and hence there will be an “infinite number” of solutions. In the
second equation, 3x + 4 = 3x + 12, which is a contradiction, so no solution is possible.]
A linear equation in more than one variable will not have a unique
solution, e.g. 3x + 4y = 2 (see Section 2 below).
5–3
5.2
Linear Equations in two Variables
Examples:
(a) 3x + 4y = 0 i.e. y = − 43 x has an “infinite number” of solutions which may be
tabulated thus:
x
···
-1
0
1
4
3
2
···
y
···
3
4
0
- 43
-1
-1 12
···
For every value of x, there is a value of y, e.g., x = 43 , y = −1.
(b) 3x + 4y = 2 i.e. y =
1
2
−
3x
4
has a infinite number of solutions, e.g.
x
···
-1
0
1
4
3
2
···
y
···
1 14
1
2
− 14
− 12
-1
···
Exercise 5.2: Solve 2x − y = 3
5.3
Pair of Linear Equations in two Variables
Suppose we are given the equations 3x + 4y = 2 and x + y = 3. Each equation separately
has an infinite number of solutions. Taken together, however, these two equations have a
unique solution.
The equations are then said to be solved simultaneously i.e., “at the same time”.
There are two equivalent methods of solution to this problem.
Example:
Solve 3x + 4y = 2 . . . . . . . . . . . . . . . . . . (i)
x+y =3
. . . . . . . . . . . . . . . . . . (ii)
)
simultaneous
linear equations
First solution:
(ii) × (4) :
4x + 4y = 12 . . . . . . . . . . . . . . . . . . (iii)
3x + 4y = 2 . . . . . . . . . . . . . . . . . . (i)
(iii) − (i) :
x
= 10
5–4
Substitute in (ii):
10 + y = 3
10 + y − 10 = 3 − 10
y = −7
x = 10
∴ the unique solution is
y = −7
Always substitute these values of x and y in (i), (ii) to check the correctness of the
solution.
Second solution: Write (i), (ii) as
3
1
y = − x + . . . . . . . . . . . . . . . (i)0
4
2
y = −x + 3 . . . . . . . . . . . . . . . . . . (ii)0
(i)0 , (ii)0 give − 43 x +
1
2
= −x + 3 (= y) which we solve as in Topic 5, Section 1.
3
1 1
− x+ −
4
2 2
3
− x
4
3
− x+x
4
1
x
4
1
x×4
4
x
Substitute for x in (i)0 : y = − 43 × 10 +
= −x + 3 −
1
2
1
2
1
−x + 2 + x
2
1
2
2
1
2 ×4
2
10
= −x + 2
=
=
=
=
1
2
= −7 12 +
1
2
= −7
0
Substitute for x in (ii) : y = −10 + 3 = −7
x = 10
∴ the unique solution is
y = −7
[The checking of the solution is inherent in the process.]
5–5
Exercises 5.3:
(i) Solve
(a) 2x + 3y = 10, 3x − 2y = −11
(b) 5x − 2y = 1, 2x = 4 − y
(ii) Why do the pairs of equations
(a) x − y = 5, 3x − 3y = 4
(b) x − y = 5, 3x − 3y = 15
not have a unique solution?
5.4
Geometrical Interpretation
The unique solution of 2 simultaneous linear equations gives the
unique point of intersection of 2 straight lines.
See the next Topic, Graphs of Straight Lines.
5–6
5.5
Answers to Exercises
5.1:
(i) 2
(iii) 2 12
(v) −3
(ii) 1 34
(iv) −2
(vi)
5.2:
(
x = . . . , −1,
0,
1
2
1, 1 12 , 2, . . .
y = . . . , −5, −3, −1,
0, 1, . . .
(vii) 1
(viii) 0
are some solutions.
5.3:
(i) (a) x = −1, y = 4
(b) x = 1, y = 2
(ii) (a) the two equations are inconsistent. (if x − y = 5 is true, then 3x − 3y =
3(x − y) = 3 × 5 = 15, not 4)
(b) the second equation is exactly 3 times the first equation, i.e., there is only one
(independent) equation, i.e., the equations are not independent.
5–7